Topic: Machining and Machine Tool Operations
Tool life of 10 hrs is obtained when cutting with single point tool at 63 m/min. If Taylor's constant C = 257.35, tool life on doubling the velocity will be
- A. 5 hrs
- B. 25.7 min
- C. 38.3 min
- D. No change
Answer: B. 25.7 min
Explanation: Given $T_1 = 10 \text{ hrs} = 600 \text{ min}$ and $V_1 = 63 \text{ m/min}$. Using Taylor's tool life equation $V_1 T_1^n = C$, we have $63 \times (600)^n = 257.35$. This gives $600^n = 4.085$, resulting in $n \approx 0.22$. When the velocity is doubled, $V_2 = 126 \text{ m/min}$. Then, $126 \times T_2^{0.22} = 257.35$, giving $T_2^{0.22} = 2.042$. Solving for $T_2$ yields approximately 25.7 minutes.